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If $f:[0,1]\to R$ be a continous function such that $f(x)\leq x^{3}$ for all $ x \in [0,1]$ with$ \int_{0}^{1}f(x)dx=\frac{1}{4}$ .Then $f(x)=x^3$for all x $\in R$.How to prove it.

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2 Answers 2

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Hint: let $g(x) = x^3 - f(x)$. Then $g(x) \geqslant 0$ for $x \in [0, 1]$ and

$$\int \limits_0^1 g(x) \, \mathrm{d} x = 0.$$

Can you prove that $g(x) = 0$ for $x \in [0, 1]$ ?

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Hint: What happens if $f(x)<x^3$ for some $x\in[0,1]$? (Remember: the function $f$ is supposed to be continuous!)

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